Java strings: program output not as expected

Question

String s="abc";
String s1=s;
s=s+"d";
System.out.println(s==s1 +" "+ s.equals(s1));
System.out.println(s.equals(s1));

The above code is written in java. I thought the output of the above program might be

false false
false

but the actual conclusion

false 
false

Can someone explain why this is the result and why not like the previous one.

Thanks in advance.

+3

Gowtham kesa 10 jul. '15 at 5:30

2 answers

Let's analyze what's going on here:

System.out.println(s==s1 +" "+ s.equals(s1));

You have boolean + string + boolean. But keep in mind that operators ==

and +

got different priorities (see. Http://bmanolov.free.fr/javaoperators.php ).

As you can see, the + operator has higher precedence and is therefore executed first. This brings up the following:

System.out.println(s == (s1 +" "+ s.equals(s1)) );

As you can see, the line s1

will be appended to " "

, and the rest is false

boolean.

So you have

System.out.println(s == "someString");

which is false and that's why you get "false"

as output.

+2

davidgiga1993 10 jul. 15 at 5:45 am

DTing · Accepted Answer · 2015-07-10T05:37:00+0000

System.out.println(s==s1 +" "+ s.equals(s1));

equivalent to:

System.out.println(s==(s1 +" "+ s.equals(s1)));

if you used:

System.out.println((s==s1) + " " + s.equals(s1));

You'll get:

false false